Lp theory for outer measures:Application to time-frequency analzysis

Christoph ThieleSantander, September 2014

Recall Tents (or Carleson boxes)

X is the open upper half plane, generating sets are tents T(x,s) :

Define outer measure on X by

€

σ(T(x,s)) := s

Sizes

Sp average of function on a tent

Also Sinfty

€

S p ( f )(T(x,s)) = (1

sf (y, t)p dy

dt

tT (x,s)

∫∫ )1/ p

€

S∞( f )(T(x,s)) = sup(y,t )∈T (x,s) f (y, t)

Outer essential supremum

Space of functions with finite out.ess.supremum

Outer essential supremum on a subset F:€

outess( f ) = sup{S( f )(E) : E ∈Σ}

€

L∞(X,μ,S)

€

outessF ( f ) = sup{S( f 1F )(E) : E ∈Σ}

Outer Lp spaces

Define super level measures

Define Lp norms

Also weak Lp (Lorentz space) €

μ( f > λ ) := inf{μ(F) : outessF c f ≤ λ}

€

fLp (X ,μ ,S )

:= ( pλ p

0

∞

∫ μ( f > λ )dλ /λ )1/ p

€

fLp ,∞ (X ,μ ,S )

:= supλ λ pμ( f > λ )

Embedding theorems

Thm: Define for fixed Schwartz function ϕ

Then we have:

If φ has integral zero:

€

Fφ ( f )(y, t) := f (z)t−1φ(t−1(z − y))dz∫

€

Fφ ( f )Lp (X ,μ ,S ∞ )

≤ Cφ fp

€

Fφ ( f )Lp (X ,μ ,S 2 )

≤ Cφ fp

Paraproduct estimates

For two phi-I with mean 0 and one with mean 1

Hoelder with two qi equal 2 and one equal infty

€

Λ( f1, f2, f3) := ( Fφ i( f i)(y, t))dy

dt

ti=1

3

∏∫∫

€

Λ( f1, f2, f3) ≤ Fφ i( f i)

i=1

3

∏L1 (X ,μ ,S1 )

≤ C Fφ i( f i)

i=1

3

∏Lpi (X ,μ ,Sqi

)

≤ C f ii=1

3

∏Lpi

Lerner’s sparse operator estimates

For two phi-I with mean 1 and some G

Hoelder with two qi equal infty and one equal 1

€

ΛG ( f1, f2) := G(y, t)( Fφ i( f i)(y, t))dy

dt

ti=1

2

∏∫∫

€

ΛG ( f1, f2) ≤ G Fφ i( f i)

i=1

2

∏L1 (X ,μ ,S1 )

≤ C GL∞ ( X , μ ,S1 )

Fφ i( f i)

Lpi ( X , μ ,S∞ )i=1

2

∏ ≤ C f ii=1

2

∏Lpi

Bilinear Hilbert transform

Consider vectorDefine the trilinear form

The bilinear Hilbert transform is any of the pre-dual bilinear operators.

May assume beta nonzero. By scaling may assume it is unit length. By change of variables may assume it is perpendicular to (1,1,1).

€

(β1,β2,β3)∈R3

€

Λ( f1, f2, f3) := p.v. ( f i(x − βit)i=1

3

∏ )dxdt

t∫∫

Degenerate case

If two beta-I are equal, degenerate case.

Reduce to pointwise product and Hilbert transf.We consider only one fixed non-degenerate case

€

Λ( f1, f2, f3) := p.v. f1(x − βt) f2(x − βt) f3(x + 2βt)dxdt

t∫∫

€

Λ( f1, f2, f3) := p.v. ( f1 f2)(x − 3βt)dt

t∫∫ f3(x)dx

Boundedness of BHT

Theorem: Fix nondegenerate direction beta. Let

Then there is an a priori estimate

Extensions exist for p-I less than 2, but that is for some other time. €

Λ( f1, f2, f3) ≤ C f i p ii=1

3

∏€

2 < p1, p2, p3 < ∞

€

1

p1

+1

p2

+1

p3

=1

Symmetries

Like CZ operators, BHT is invariant under simultaneous translation of f-i and has proper scaling under simultaneous dilation of f-i.

It has additional one parameter modulation symm.

This equals original BHT if alpha perpendicular to beta and (1,1,1). BHT shares symmetries with

€

Mη f (x) = e ixη f (x)

€

Λ(Mα 1η f1,Mα 1η f2,Mα 1η f3) := p.v. ( f i(x − βit)ei(x −β i t )α iη

i=1

3

∏ )dxdt

t∫∫

€

˜ Λ ( f1, f2, f3) = f1(x)R

∫ f2(x) f3(x)dx

New embedding map

Additional symmetry makes the upper half plane embedding inappropriate and useless.

Need embedding map reflecting the orbit of entire symmetry group :

Phi assumed to have FT nonnegative and with small support near 0. The testing wavelets become arbitrarily high oscillating wavepacket. €

Fφ (y,η, t) = f (z)e izη∫ t−1φ(t−1(z − y))dz

BHT in terms of embedded fcts

(Remark: The RHS is the bilinear Cauchy projection.)Two ways of proving this algebraic identity:1) Exercise in applying Plancherel, akin to the proof of

the Calderon reproducing formula.2) Checking RHS satisfies the symmetries, and noting

that any form with these symmetries is a linear combination as on LHS. Prove RHS not multiple of Lambda tilde to show a not 0.

€

aΛ( f1, f2, f3) + b ˜ Λ ( f1, f2, f3) = F j (y,α jη + β j t−1, t)dη dydt

j =1

3

∏R

∫R

∫0

∞

∫

Proof of symmetries

By change of variables.

Translation symmetry: y to y+y’Modulation symmetry: eta to eta+eta’Dilation symmetry: t to ts, y to ys, eta to eta/s

€

F j (y,α jη + β j t−1, t)dηdydt

j =1

3

∏R

∫R

∫0

∞

∫

€

F j (y,η, t) = f j (z)e izη∫ t−1φ(t−1(z − y))dz

€

Not multiple of Lambda tilde

For fixed eta,y,t we have a pairing of with a certain nonnegative bump.

This bump is supported on the half space If is supported outside this half space,

then form vanishes. If supported and positive in half space, then form does not vanish.

So form not invariant under beta-modulation.

€

F j (y,α jη + β j t−1, t)dηdydt

j =1

3

∏R

∫R

∫0

∞

∫

€

F(y,η, t) = f (z)e izη∫ t−1φ(t−1(z − y))dz

€

€

ς⋅β > 0

€

ˆ f 1(ς1)ˆ f 2(ς 2) ˆ f 3(ς 3)

€

ˆ f 1(ς1)ˆ f 2(ς 2) ˆ f 3(ς 3)

Define outer measure on upper 3space

€

ξ

€

T(x,ξ,s) = {(y,η, t) : y − x < s − t,|η − ξ |< t−1}

€

η

€

σ(T(x,ξ,s)) = s

€

S(F)(T(x,ξ,s)) =1

sF(y,η, t)

T (x,ξ ,s)

∫∫∫ dydη dt

Outer triangle inequality

where

€

F j (y,α jη + β j t−1, t)dηdydt

j =1

3

∏R

∫R

∫0

∞

∫

€

= G j (y,η, t)dη dydtj =1

3

∏R

∫R

∫0

∞

∫

€

≤ G1G2G3 L1 (X ,μ ,S )

€

G j (y,η, t) = F j (y,α jη + β j t−1, t)

Outer Hoelder inequality

Where we need to identify appropriate sizesS1,S2,S3 which satisfy size-Hoelder€

G1G2G3 L1 (X ,μ ,S )≤ C G1 Lp1 (X ,μ ,S1 )

G2 Lp2 (X ,μ ,S2 )G3 Lp3 (X ,μ ,S3 )

€

S(G1G2G3)(T) ≤ S1(G1)(T)S2(G2)(T)S3(G3)(T)

Idea of sizes

The j-size is a sum of L2 and Linfty average. The L2 is taken outside an exceptional region A-j. The three exceptional regions are disjoint.

€

S j (G)(T(x,ξ,s)) =1

sG(y,η, t)

2

T (x,ξ ,s)\ A j

∫∫∫ dydηdt ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

1/ 2

+ sup(y,η ,s)∈T (x,ξ ,s) G(y,η, t)

Size-Hoelder

To proveWe split the integral over tent on LHS into four

regions: three orange regions and the rest.On each exceptional region we do the

appropriate 2-2-infty Hoelder, since we control two functions in L2. On the rest, either of the Hoelders is OK.

€

S(G1G2G3)(T) ≤ S1(G1)(T)S2(G2)(T)S3(G3)(T)

Modified outer measure space

To formulate sizes precisely, express everything for F rather than G (change variables). Fix j.

Tents:

Size (pick small parameter b): Note: regions AWill be disjoint for different j after change vrbls.€

T(x,ξ,s) = {(y,η, t) : y − x < s − t,| α (η − ξ) + βt−1 |< t−1}

€

σ(T(x,ξ,s)) = s

€

S(F)(T(x,ξ,s)) =1

sF(y,η, t)

2

T (x,ξ ,s)\ A

∫∫∫ dydηdt ⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2

+ sup(y,η ,s)∈T (x,ξ ,s) F(y,η, t)

€

A = {(y,η, t) : y − x < s − t,|η − ξ |< bt−1}

Main embedding theorem

After this preparation, bounds for BHT are reduced to embedding theorem

For 2<p<infty

€

Fφ Lp (X ,μ ,S )≤ C f

p

Few comments on the proof

Story told by pictures…

ξ

x